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STRUCTURE OF O-16 AND O-15
By Prof. Lefteris Kaliambos (Λευτέρης Καλιαμπός) Τ.Ε. Institute of Larissa Greece. ( June 2014) Historically the discovery of the assumed uncharged neutron (1932) along with the invalid relativity (EXPERIMENTS REJECT RELATIVITY) led to the abandonment of the well-established electromagnetic laws, in favour of various contradicting nuclear theories, which could not lead to the nuclear structure. Under this physics crisis and using the charged UP and DOWN quarks , discovered by Gell-Mann and Zweig, I published my paper “Nuclear structure is governed by the fundamental laws of electromagnetism ” (2003), which led to my discovery of the new structure of protons and neutrons given by proton = + 5d + 4u = 288 quarks = mass of 1836.15 electrons neutron = + 4u + 8d = 288 quarks = mass of 1838.68 electrons The paper was also presented at a nuclear conference held at NCSR "Demokritos" (2002). In this photo I present the electromagnetic laws governing the nuclear structure of magic nuclei, but a student of Einstein (Dr Th. Kalogeropoulos ) criticised my discovery of nuclear force and structure by believing that the nuclear structure is due to the invalid relativity In fact, here one can see the 9 charged quarks in proton and the 12 ones in neutron able to give the charge distributions in nucleons for revealing the strong electromagnetic force for the nuclear binding in the correct nuclear structure by applying the laws of electromagnetism. You can see my papers of nuclear structure in my FUNDAMENTAL PHYSICS CONCEPTS . Note that according to my discovery of the LAW OF ENERGY AND MASS the mass defect in the nuclear structure is due to the photon mass of the emitting dipolic photon presented at the international conference "Frontiers of fundamental physics" (1993) organised by the natural philosophers M. Barone and F. Selleri , who gave me an award including a disc of the atomic philosopher Democritus. Nevertheless today many physicist continue to apply not the well-established laws but the various fallacious nuclear structure models which lead to complications. ' ' Nuclear structure of Oxygen -16 with S = 0 To compare the structure of O-16 with the structures of He-4, Be-8, and Pb-208 see the following figures. Also in the following diagram of the structure of O-16 one sees that the symmetry and the total spin are the two fundamental characteristics which lead to the nuclear structure of O-16. In Fig. 6b of my published paper you see that the structure of O-16 consists of four horizontal squares. The nucleons of the second and third square form a parallelepiped of a great stability, because it is characterized by points with four pn bonds per nucleon. This is the central parallelepiped of the structure of O-16 which is responsible for the stability of O-16. In the following diagram you see also that p1, p2, n1 and n2 form the first horizontal square, while the p3, p4, n3 and n4 form the second horizontal square. Such nucleons along with the p5, p6, n5 and n6 of the third square form the very strong central parallelepiped which we call central Be8 with a binding energy B(central Be8) > B(Be8). The structure of this is similar to the isolated Be8 but it is very stable because it has four pn bonds per nucleon. For example at point p3 we observe the four pn bonds as p3n1, p3n3, p3n4 and p3n5. Because of this condition we assume that its binding energy has approximately the half value of the binding energy of O-16 That is B( Central Be8 ) = B(O-16)/2 = -127.58/2 = -64 MeV Whereas the binding energy of the unstable Be8 is B(Be8 = - 56.5 MeV. In other words such a binding energy of the central Be8 is responsible for the stability of O-16 which is the magic nucleus in the group of the simple parallelepipeds called alpha particle nuclei. Structure of O-16 with S = 0 p8(-1/2).n8(-1/2) n7(-1/2).p7(-1/2) Fourth horizontal square of negative spins n6(+1/2).p6(+1/2) p5(+1/2).n5(+1/2) Third horizontal square of positive spins p4(-1/2).n4(-1/2) n3 (-1/2).p3 (- 1/2). Second horizontal square of negative spins n2(+1/2).p2(+1/2). p1(+1/2). n1(+1/2) First horizontal square of positive spins Decay of Oxygen-15 leads to the structure of N-15 Looking carefully at the above diagram one sees that the structure of O-15 is due to the absence of n1(+1/2) or n2(+1/2). Thus, the total spin of the O-15 is S = -1/2 . For example in the structure of O-16 adding the spins of nucleons of the four squares along the same z axis we get S = n1(+1/2) + p3(-1/2) + (n5(+1/2) + p7(-1/2 = 0 Whereas in the structure of O-15 because of the absence of n1 we write S = p3(-1/2) + n5(+1/2) + p7(-1/2) = -1/2 In this case since the symmetry brakes, the binding energy from B(O-16) = -127.58 MeV reduces to B(O-15) = - 119.67 MeV. Of course it is due to the fact that the absence of n1 with three np bonds at point n1 reduces the binding energy. Under this condition also the p1 has now only the two p1n2 and p1n3 bonds, while in the structure of O-16 it had the three pn bonds per nucleon like the p1n1 the p1n2 and the p1n3. This situation with only two pn bonds is not favorable energetically, because all protons of the new structure exert repulsions on p1, So this proton cannot stay there because of the only two pn bonds. Therefore according to the following reaction the p1under a repulsive energy Q due to all rest protons turns to a new neutron n . p1 + Q = n + positron + neutrino At the same time the new neutron (n) goes at point n1 in order to make the bonds np2 and np3. This situation of two np bonds per neutron replaces both n1 and p1 because the braking of symmetry leads to a new structure in which the repulsive energy of the pp systems is reduced. Such a structure of 7 protons and 8 neutrons gives the structure of stable N-15. In other words N-15 has a core just the structure of O-16 in which a proton is absent for breaking the symmetry of O-16. Category:Fundamental physics concepts